package com.zrf.base.knowledge.leetcode.editor.cn;

import java.util.ArrayList;
import java.util.List;

/**
 * //给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。
 * //
 * // 给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
 * //
 * //
 * //
 * // 示例:
 * //
 * // 输入："23"
 * //输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
 * //
 * //
 * // 说明:
 * //尽管上面的答案是按字典序排列的，但是你可以任意选择答案输出的顺序。
 * // Related Topics 字符串 回溯算法
 * // 👍 966 👎 0
 *
 * @author zhouruifeng
 * @date 2020-10-24 16:05:55
 */
public class LetterCombinationsOfAPhoneNumber {
    public static void main(String[] args) {
        Solution solution = new LetterCombinationsOfAPhoneNumber().new Solution();
        System.out.println(solution.letterCombinations("23"));
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        private String[] phoneResources = new String[]{"", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

        public List<String> letterCombinations(String digits) {
            List<String> resultList = new ArrayList<>();
            if ("".equals(digits)) {
                return resultList;
            }

            combinations(digits, resultList, new StringBuilder());
            return resultList;
        }

        private void combinations(String digits, List<String> resultList, StringBuilder tmp) {
            if (digits.length() == tmp.length()) {
                resultList.add(tmp.toString());
            } else {
                int index = tmp.length();
                String phoneResource = phoneResources[Integer.parseInt(digits.substring(index, index + 1)) - 1];
                for (int i = 0; i < phoneResource.toCharArray().length; i++) {
                    tmp.append(phoneResource.charAt(i));
                    combinations(digits, resultList, tmp);
                    tmp.deleteCharAt(tmp.length() - 1);
                }

            }
        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}